Modul 2 - Konservasi Energy Bernoully

8/4/2019 Modul 2 - Konservasi Energy Bernoully

1/57

Konservasi Energi

(Kontinuitas & Hk. Bernoulli)


2/57

Mass Flux

Tube of flow: bundle of streamlines

A1 A2

1vP

2v

Q

11 1 1 1 1 1 1

1

mass fluxm

m Av t Avt


3/57

Persamaan Kontinuitas

Gambar Aliran masuk dan keluar dari bagian pipa arus

Volume aliran masuk dititik 1 :

A1

. v1

. dt Dimana : A1

= luas penampang dan v1

. dt = tinggi saluran

Volume aliran masuk dititik 2 :

A2. v2. dt Dimana : A2 = luas penampang dan v2. dt = tinggi saluran

Jika kerapatan fluida (densitas) =

Maka massa aliran : . A1. v1 dan . A2. v2Avdtm ..


4/57

Conservation of Mass

IF: no sources and no sinks/drains

1 1 1 2 2 2

1 1 2 2

constant

co fornstant incompressible fluid,

Av A v

Av A v

Example of equation of continuity. Also conservation ofcharge in E&M

Narrower tube = larger speed, fastWider tube = smaller speed, slow

1 = 2

A x V = constan

Hukum kontinuitas

Q = A x V = konstans


5/57

Massa fluida yang mengalir adalah konstan disepanjang saluran, maka :

. A1. v1 = . A2. v2

atau . A . v = konstan Hukum kekekalan massa aliran

Dengan massa aliran dengan massa jenis tetap : , maka :

A1. v1 = A2. v2

A . v = konstanatau Hukum kekekalan volume aliran

Jika : Q = Kapasitas atau jumlah aliran (Debit) (m3/detik)

Q = A . v= konstan Persamaan Kontinuitas

Dimana : A = Luas penampang (m2) dan v= kecepatan aliran (m/detik)


6/57

FLUID DYNAMICSTHE BERNOULLI EQUATION

The laws of Statics that we have learned cannot solveDynamic Problems. There is no way to solve for the flowrate, or Q. Therefore, we need a new dynamic approach

to Fluid Mechanics.


7/57

Example of equation of continuity.

Penampang 1 sebuah saluran yang mengalirkan air dengan kecepatan 3 m/sdan garis tengahnya 2 m, dipenampang 2 garis tengahnya 3 m, berapakahDebit atau kapasitas dan kecepatan dipenampang keluar (2)

1V1

2V2

Penyelesaian :

Persamaan kontinuitas :

Q = A x V = konstans

Q1= A1x V1 Q2= A2x V2&

Q1= Q2=Q

Q=Q2= A2x V2

s

mVdQQ

32

1

2

11 42.93.)2(4

.4

sm

d

x

A

QV /33.1

.

42.94

222

2


8/57

Conservation of Energy

Steady, incompressible, nonviscous, irrotational


9/57

Bernoullis Principle


10/57


Flow is faster when the pipe is narrower Put your thumb over the end of a garden

hose Energy conservation requires that the

pressure be lower in a gas that is moving

faster Has to do with the work necessary to

compress a gas (PV is energy, more later)


11/57


When the speed of a fluid increases,internal pressure in the fluid decreases.


12/57



13/57


Why the streamlines are compressed isquite complicated and relates to the air

boundary layer, friction and turbulence.


14/57



15/57

Who Accelerates the Fluid?

Acceleration due to pressure difference.Bernoullis Principle = Conservation of energy


16/57

Bernoullis Equation

1 1 2 2

2 2

1 1 1 2 2 2 2 2 1 1

2 2

1 1 1 2 2 2

2

1 1

2 2

1 12 2

1constant

2

m A x A x

p A x p A x mv mgy mv mgy

p v gy p v gy

p v gy

kinetic E, potential E, external work

Bentuk dan ukuran yang sama

Energi Balance :


17/57

Subskrip 1 dan 2 pada sembarang titik, maka ditulis menjadi :

tan

2

1 2konsvgyp

Persamaan Bernoulli

Jika : = Berat jenisg.

Maka : tan

2

1 2konsv

gy

p

Atau :

tankonsyg

vp

2

2

Energi Tekanan

Energi Kinetik

Elevasi(Energi Potensial)


18/57

The Bernoulli Equation (unit of L)

At any two points on a streamline:

P1/ + V12/2g + z1= P2/ + V2

2/2g + z2

12


19/57

The Bernoulli Equation (unit of L)

At any two points on a streamline:

P1/ + V12/2g + z1= P2/ + V2

2/2g + z2

12


20/57

A Simple Bernoulli Example

V2Z = air

Determine the difference in pressure between points 1 and 2

Assume a coordinate system fixed to the bike (from this system,the bicycle is stationary, and the world moves past it). Therefore,

the air is moving at the speed of the bicycle. Thus, V2 = Velocity ofthe Biker

Hint: Point 1 is called a stagnation point, because the air particlealong that streamline, when it hits the bikers face, has a zero

velocity (see next slide)


21/57

Stagnation Points

On any body in a flowing fluid, there is a stagnation point. Some fluidflows over and some under the body. The dividing line (the stagnationstreamline) terminates at the stagnation point. The Velocity decreases

as the fluid approaches the stagnation point. The pressure at thestagnation point is the pressure obtained when a flowing fluid is

decelerated to zero speed by a frictionless process


22/57

Apply Bernoulli from 1 to 2

V2Z

Point 1 = Point 2

P1/air+ V12/2g + z1 = P2/air+ V22/2g + z2

Knowing the z1 = z2 and that V1= 0, we can simplifythe equation

P1/air= P2/air+ V22/2g

P1 P2= ( V22

/2g ) air

= air

A Si l B lli E l


23/57

A Simple Bernoulli Example

If Lance Armstrong is traveling at 20 ft/s, what pressure

does he feel on his face if the air= .0765 lbs/ft3?

We can assume P2 = 0 because it is only atmospheric pressure

P1 = ( V22/2g )(air) = P1 = ((20 ft/s)

2/(2(32.2 ft/s2)) x .0765 lbs/ft3

P1 =.475 lbs/ft2

Converting to lbs/in2 (psi)

P1 = .0033 psi (gage pressure)

If the bikers face has a surface area of 60 inches

He feels a force of .0033 x 60 = .198 lbs

B lli A i


24/57

Bernoulli Assumptions

Key Assumption # 1

Velocity = 0

Imagine a swimming pool with a small 1 cm hole on the floor of

the pool. If you apply the Bernoulli equation at the surface, andat the hole, we assume that the volume exiting through the holeis trivial compared to the total volume of the pool, and thereforetheVelocity of a water particle at the surface can be assumed tobe zero

There are three main variables in the Bernoulli EquationPressure Velocity Elevation

To simplify problems, assumptions are often made toeliminate one or more variables

B lli A i


25/57


Key Assumption # 2

Pressure = 0

Whenever the only pressure acting on a point isthe standard atmospheric pressure, then thepressure at that point can be assumed to be zerobecause every point in the system is subject tothat same pressure. Therefore, for any free

surface or free jet, pressure at that point can beassumed to be zero.

B lli A ti


26/57


Key Assumption # 3

The Continuity Equation

In cases where one or both of theprevious assumptions do not apply, then

we might need to use the continuityequation to solve the problem

A1V1=A2V2

Which satisfies that inflow and outfloware equal at any section

B lli E l P bl F J t


27/57

Bernoulli Example Problem: Free Jets

What is the Flow Rate at point 2? What is the velocity at point 3?

1

2

3

H2O

Part 1:Apply Bernoullis eqn between points 1 and 2

P1/H2O + V12/2g + h = P2/H20 + V2

2/2g + 0

simplifies to

h = V22/2g solving for V

V = (2gh)

Q = VA or Q = A2(2gh)0A2

Givens and Assumptions:Because the tank is so large, we assume V1 = 0 (Volout


28/57

1

2

3

H2O

Z = 0A2

Bernoulli Example Problem: Free Jets

Part 2: Find V3?

Apply Bernoullis eq from pt 1 to pt 3

P1/H2O + V12

/2g + h = P3/H20 + V32

/2g HSimplify to h + H = V3

2/2g

Solving for V V3 =( 2g ( h + H ))

Th C ti it E ti


29/57

The Continuity EquationWhy does a hose with a nozzle shoot water further?

Conservation of Mass:In a confined system, all of the mass that enters the system, must also exit

the system at the same time.

Flow rate = Q = Area x Velocity

1A1V1(mass inflow rate) = 2A2V2( mass outflow rate)

If the fluid at both points is thesame, then the density drops

out, and you get the continuityequation:

A1V1 =A2V2

ThereforeIf A2 < A1 then V2 >V1

Thus, water exiting a nozzle has

a higher velocity

Q1 = A1V1

A1

V1 ->

Q2 = A2V2

A1V1 = A2V2

A2 V2 ->


30/57

Free Jets

The velocity of a jet of water is clearly related to the depth of waterabove the hole. The greater the depth, the higher the velocity. Similarbehavior can be seen as water flows at a very high velocity from the

reservoir behind a large dam such as Hoover Dam


31/57

example problem

A tank of water has a small nozzle at itsbase as shown. Find the velocity inft/sec and the volumetric flow rate in

ft

3

/sec from the nozzle.


32/57

Fluid Mechanics

Assume the jet is cylindrical and that thepressure is atmospheric as soon as itleaves the nozzle. Apply Bernoullis

equation between a point 1 on the surfaceof the tank and a point 2 at the nozzleexit:

2

2

221

2

11 zg2g

Pzg2g

P

vv


33/57

Fluid Mechanics

The pressure at point 1 and at point 2 isjust that of the atmosphere, so P1 = P2.

At point 1 the height is z1 = H and at point2 the height is z2 = 0. If the size of thetop of the tank is very large compared to

the outlet area, then (V1)2 is significantlyless than (V2)

2 .


34/57

Fluid Mechanics

The Bernoulli Equation becomes

but the terms P1/g and P2/g are equaland cancel because the pressures are thesame, and what is left is

0g2g

P

H0g

P2

221

v

gH22 V


35/57

Fluid Mechanics

And so the exit velocity is

the discharge flow rate is

sec/ft1.32)ft16)(sec/ft2.32(2 22

V

3in ft ft

in sec sec

ft

2

2

2 2 2

41 132.1 2.8

4 412

Q A d

V V

Tank Example


36/57

Tank ExampleSolve for the Pressure Head, Velocity Head, and Elevation Head at each

point, and then plot the Energy Line and the Hydraulic Grade Line

1

23 4

1

4

Assumptions and Hints:

P1 and P4 = 0 --- V3 = V4 same diameter tube

We must work backwards to solve this problem

R = .5

R = .25


37/57

1

23 4

1

4

Point 1:

Pressure Head : Only atmospheric

P1/ = 0Velocity Head : In a large tank,V1 = 0 V12/2g = 0

Elevation Head : Z1= 4

R = .5R = .25


38/57

1

23 4

1

4

H2O= 62.4 lbs/ft3

Point 4:

Apply the Bernoulli equation between 1 and 40 + 0 + 4 = 0 + V4

2/2(32.2) + 1

V4 = 13.9 ft/s

Pressure Head : Only atmospheric P4/ = 0Velocity Head :V4

2/2g = 3


R = .5R = .25


39/57

1

23 4

1

4

Point 3:

Apply the Bernoulli equation between 3 and 4 (V3=V4)P3/62.4 + 3 + 1 = 0 + 3 + 1

P3 = 0

Pressure Head : P3/ = 0Velocity Head :V3

2/2g = 3


R = .5R = .25


40/57

1

2 3 4

1

4

Point 2:

Apply the Bernoulli equation between 2 and 3P2/62.4 + V2

2/2(32.2) + 1 = 0 + 3 + 1

Apply the Continuity Equation

(P.52)V2 = (P.252)x13.9 V2 = 3.475 ft/s

P2/62.4 + 3.4752/2(32.2) + 1 = 4 P2 = 175.5 lbs/ft

2

R = .5R = .25

Pressure Head :P2/ = 2.81

Velocity Head :V2

2/2g = .19

Elevation Head :Z2= 1

Penerapan Persamaan Bernoulli


41/57

Dengan mengambil dasar tangki sebagai patokan, diperoleh :

2

2

2

12

1

2

1vpghvp a

Atau gh

ppvv a 22

2

1

2

2

Berdasarkan hukum kontinuitas :

1

2

12 v

A

Av

Disebut vena contracta

Jika tangki terbuka dan berhubungan dengan udara luar, maka : p = pa dan p pa = 0

Dan : A1 >> A2, maka : ghv 22 Kecepatan efflux sama dengan kecepatan benda jatuh bebas melaluiTinggi h disebut Dalil Toricelli

Jika pada permukaan tertutup dimana v12 diabaikan dan tekanan p besar maka :


42/57

ppv a

)(22

Penerapan pada gaya dorong sebuah roket

Jika : A = luas lubang, = rapat massa fluida yang keluarv = kecepatan efflux,

maka massa fluida yang mengalir dalam waktu dt :

A v dt

maka momentumnya : A v2 dtkarena kecepatan fluida didalam relatif kecil, maka : A v2Maka gaya reaksi dorongan :

Atau :

ppAAvF o

)(22

)(2 appAF

Gambar 1

Contoh : (gambar 1)


43/57

Sebuah roket dimana disebelah atas air terdapat udara bertekanan p = 2 atm.a). Jika roket ditahan diam, berapakah kecepatan efflux yang keluar dari lubang pada ekor roketb). Berapa tolakan keatas (dorongan) keatas jika luas lubang 0,5 cm2 ?

a. Tekanan relatif p pa = 1 atm 106 dyn cm-2, maka kecepatannya :

ppv a

)(22

3

25

.1

.102

cmgr

cmdynexv

b. Gaya tolak keatas (dorongan) atau gaya reaksi :

dyncmdynxcmxF6262

10.105,02

Pipa Ventury


44/57

2

2

221

2

112

1

2

1gyvpgyvp

Dari persamaan Bernoulli :

y1 = y2

2

22

2

112

1

2

1

vpvp Maka :

Berdasarkan persamaan kontinuitas kecepatan dititik 2 lebih besar dititik 1, maka sebaliknyatekanan dititik 2 (tenggorokan) lebih kecil dari tekanan dititik 1, dengan demikian gaya nettomenuju kekanan memberi percepatan pada fluida ketika memasuki tenggorokan, maka untukmengukur kecepatan dan massa yang mengalir digunakan Ventury meter

Mengukur tekanan didalam fluida yang mengalir


45/57

p adalah tekanan statik fluida yang mengalir didalam saluran tertutup dan jika perbedaan tinggi zatcair h1 yang besarnya adalah sebanding dengan perbedaan tekanan atmosfir pa dan tekanan statikp yaitu :

pa p =mg h1p =mg h1

Dimana : m= rapat massa zat cair


46/57

BEq in Everyday Life

Open a faucet, thestream of water gets

narrower as it falls.

Velocity increases due to

gravity as water flow down,thus, the area must get narrower.

A1

A2

V1

V2


47/57

Q & A on Bernoullis Eq.

A bucket full of water.

One hole andone pipe, bothopen at bottom.

Out of which water flows faster?

Same. It only depends on depth.


48/57

Measuring Pressure

E. Torricelli: Mercury Barometer

hpatm

p=0atm

atm

P ghP

hg


49/57

U-Tube Manometer

1 1 2 2A atmp gh p gh


50/57

The Venturi Meter

Speed changes asdiameter changes.

Can be used tomeasure the speedof the fluid flow.

2 2

1 1 2 2 1 1 2 21 1 ,2 2

p v p v v A v A


51/57

The Pitot Tube

21

2a a b

b a

p v p

p p gh

Th E Li d th H d li G d Li


52/57

The Energy Line and the Hydraulic Grade Line

Looking at the Bernoulli equation again:

P/ + V2/2g + z = constant on a streamlineThis constant is called the total head (energy), H

Because energy is assumed to be conserved, at any point alongthe streamline, the total head is always constant

Each term in the Bernoulli equation is a type of head.

P/ = Pressure Head

V2/2g = Velocity Head

Z = elevation head

These three heads summed equals H = total energy

Next we will look at this graphically



53/57


Q

Measures thestatic pressure Pitot measuresthe total head

1

Z

P/

V2

/2g

EL

HGL

2

1: Static Pressure TapMeasures the sum of theelevation head and the

pressure Head.

2: Pitot Tube

Measures the Total Head

EL : Energy Line

Total Head along a systemHGL : Hydraulic Grade line

Sum of the elevation andthe pressure heads along a

system



54/57


Q

Z

P/

V2

/2g

EL

HGL

Understanding the graphical approach ofEnergy Line and the Hydraulic Grade line is

key to understanding what forces aresupplying the energy that water holds.

V2/2g

P/

Z

1

2

Point 1:

Majority of energystored in the water is in

the Pressure Head

Point 2:

Majority of energystored in the water is inthe elevation head

If the tube wassymmetrical, then thevelocity would beconstant, and the HGLwould be level

Plotting the EL and HGL


55/57

Plotting the EL and HGLEnergy Line = Sum of the Pressure, Velocity and Elevation heads

Hydraulic Grade Line = Sum of the Pressure and Velocity heads

EL

HGL

Z=1Z=1Z=1

V2/2g=3V2/2g=3

Z=4

P/ =2.81

V2/2g=.19


56/57


57/57

Modul 2 - Konservasi Energy Bernoully

Documents

Transcript of Modul 2 - Konservasi Energy Bernoully